【力扣刷题】17. 电话号码的字母组合

输入：digits = "23"
输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]

输入：digits = ""
输出：[]

输入：digits = "2"
输出：["a","b","c"]

class Solution {
    public List<String> letterCombinations(String digits) {
        if(digits==null || digits.length()==0) {
            return new ArrayList<String>();
        }
        //一个映射表，第二个位置是"abc“,第三个位置是"def"。。。
        //这里也可以用map，用数组可以更节省点内存
        String[] letter_map = {
                " ","*","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"
        };
        List<String> res = new ArrayList<>();
        //先往队列中加入一个空字符
        res.add("");
        for(int i=0;i<digits.length();i++) {
            //由当前遍历到的字符，取字典表中查找对应的字符串
            String letters = letter_map[digits.charAt(i)-'0'];
            int size = res.size();
            //计算出队列长度后，将队列中的每个元素挨个拿出来
            for(int j=0;j<size;j++) {
                //每次都从队列中拿出第一个元素
                String tmp = res.remove(0);
                //然后跟"def"这样的字符串拼接，并再次放到队列中
                for(int k=0;k<letters.length();k++) {
                    res.add(tmp+letters.charAt(k));
                }
            }
        }
        return res;
    }
}